Then it becomes a lot of distributing and factoring, until you can finally divide out the h. Once that happens, substitute 0 for h. Phew. I dunno if I can type out everything I did. . . I probably did it the long way, but I saw no other, shorter way. Do you want me to type it out?

Before that, all I did was find the common denominator of 8/sqrt[x+h] and 8/sqrt[x], add them together, and instead of dividing by h, multiply by 1/h to get
\[8\sqrt{x} - 8\sqrt{x+h} / h \sqrt{x} \sqrt{x+h}\]
Then I factored out the 8 to get the above ^
Oh. Okay.

I really think you are going about this the wrong way. You should change the square root into a power (1/2) and then bring the denominator to the top by changing the power to negative. Then you can simply use the limit process of:
f(x)= 8x^(-1/2)
f(x+h)=8(x+h)^(1/2)
And plug this into the derivative formula.
You should get rid of the terms in the denominator.

\[f(x)=\frac{8}{\sqrt{x}}\]
\[f(x)=\frac{8}{x ^{1/2}}\]
\[f(x)=8x ^{\frac{-1}{2}}\]
\[f(x+h)=8(x+h)^{-1/2}\]
\[f \prime(x)\lim_{h \rightarrow 0}=I can't remember the formula\]
Now just use you derivative formula, cancel where necisary, factor out a h, then cancel with the denominator.

All terms on the right hand side of the minus sign in your derivative formula that have a x term should cancel with the terms. I think you're going about this the wrong way. Instead of having a quotient you should really bring the denominator to the top by making the power negative. Remember this rule?
am^-x = a/m^x